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Q.
Area of the region bounded by the curve $y=|x+1|+1$,
$x=-3, x=3$ and $y=0$ is
Application of Integrals
Solution:
Given equation of the curves are
$y=|x+1|+1=\begin{cases}(x+1)+1, \text { if } x+1 \geq 0 \\ -(x+1)+1, \text { if } x+1<0\end{cases}$
$=\begin{cases}x+2, & \text { if } & x \geq-1 \\ -x . & \text { if } & x<-1\end{cases}\,\,\,\, \ldots .(1)$
$x=-3\,\,\,\,\,\,\,\,...(ii)$
$x=3\,\,\,\,\,\,\,\,...(iii)$
and $y=0\,\,\,\,\,\,\,...(iv)$
eq. (ii) represents the line parallel to y-axis and passes through the point $(-3,0)$.
eq. (iii) represents the line parallel to y-axis and passes through the point $(3,0)$
$\therefore $ Required area $=\int\limits_{-3}^{-1} y \,dx +\int\limits_{-1}^{3} y \,dx$
$=\int\limits_{-3}^{-1}-x\, d x+\int\limits_{-1}^{3}(x+2) d x$
$=\left[\frac{-x^{2}}{2}\right]_{-3}^{-1}+\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{3}=16$ sq. units
